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Probability Theory September 27, 2012 1 Basic Probability Definition: Algebra If S is a set then a collection A of subsets of S is an algebra if • S∈A • A ∈ A =⇒ Ac ∈ A Sn • {Ai }ni=1 ∈ A =⇒ i=1 Ai ∈ A Definition: σ-Algebra If S is a set then a collection F of subsets of S is an algebra if • S∈F • A ∈ F =⇒ Ac ∈ F S∞ • {Ai }∞ i=1 ∈ F =⇒ i=1 Ai ∈ F Lemma: The following are basic properties of σ-algebras • φ∈F • {Ai }∞ i=1 ∈ F =⇒ T∞ i=1 Ai ∈ F Lemma: If S is a non-empty set and G ∈ P(S) then ∃F := σ(G) the smallest σ-algebra containing G Proof: We know that P(S) is a σ-algebra containing G Define Γ := T {A σ-algebra : G ∈ A} Claim F = A∈Γ A If this is a σ-algebra it is clearly the smallest one containing G by definition of intersection. • S ∈ A ∀AT∈ Γ since these are σ-algebras hence S ∈ A∈Γ A • Let A ∈ F then A ∈ A ∀A ∈ Γ hence Ac ∈ A ∀A ∈ Γ hence Ac ∈ F ∞ ∞ • Let {AS i }i=1 ∈ F then {Ai }i=1 ∈ A ∞ hence Si=1 Ai ∈ A ∀A ∈ Γ ∞ hence i=1 Ai ∈ F ∀A ∈ Γ 1 1 BASIC PROBABILITY Remark: From the above lemma we say that F is the σ-algebra generated by G and that G is the generator of F Definition: BorelQσ-Algebra n Let S = Rn , G := { i=1 [ai , bi ] : ai ≤ bi ∈ Q} Then the Borel σ-algebra is B(Rn ) = σ(G) Remark: If S is a topological space then B(S) is the σ-algebra generated by open sets of S Definition: Measurable Space (S, F) is a measurable space where • S is a set • F is a σ-algebra Definition: Measurable Set A is measurable if A ∈ F where F is a σ-algebra Definition: Additive If S is a set, A is an algebra then µ : A → R|[0,∞] is additive if F, G ∈ A, F ∩ G = φ =⇒ µ(F ∪ G) = µ(F ) + µ(G) Definition: σ-Additive If S is a set, A is an algebra then µ : A → R|[0,∞] S P ∞ ∞ is σ-additive if {Ai }∞ i=1 ∈ A, Ai ∩ Aj = φ ∀i 6= j =⇒ µ i=1 Ai = i=1 µ(Ai ) Definition: Measure If (S, F) is a measurable space and µ : F → R|[0,∞] is a measure if µ is σ-additive Definition: Finite Measure A measure µ on measurable space (S, F) is finite if µ(S) < ∞ Definition: σ-Finite Measure A measure µ on measurable space (S, F) is σ-finite if ∃{Ai }∞ i=1 ∈ F s.t. • µ(Ai ) < ∞ ∀i S∞ • i=1 Ai = S Definition: Probability Measure A measure µ on S is a probability measure if µ(S) = 1 Definition: Measure Space A measure space is (S, F, µ) where S is a set, F is a σ-algebra on S and µ is a measure on (S, F) Definition: Probability Space A measure space (S, F, µ) is a probability space if µ is a probability measure Definition: π-System A family Y of subsets of set S is a π-system if Y is closed under intersection i.e. A, B ∈ Y =⇒ A ∩ B ∈ Y Definition: Dynkin System If S is a set and D is a set of subsets of S then D is a Dynkin system of S if • S∈D • A, B ∈ D, A ⊆ B =⇒ B \ A ∈ D • {Ai }∞ i=1 ∈ D, Ai ∩ Aj = φ ∀i 6= j =⇒ S∞ i=1 Ai ∈ D Remark: We denote d(G) to be the smallest Dynkin system containing G which exists Lemma: If G is a π-system then σ(G) = d(G) 2 1 BASIC PROBABILITY Proof: Clearly every σ-algebra is also a Dynkin system hence d(G) ⊆ σ(G) is trivial hence it remains to show that d(G) is a σ-algebra • Clearly S ∈ d(G) by definition • Suppose A ∈ d(G) then B \ A ∈ d(G) ∀B ∈ d(G) hence Ac = S \ A ∈ d(G) • Want to show that d(G) is a π-system define D1 := {A ⊂ S : A ∩ B ∈ d(G) ∀B ∈ G} Is a Dynkin system and G ⊆ D1 hence d(G) ⊆ D1 define D2 := {A ⊂ S : A ∩ B ∈ d(G) ∀B ∈ d(G)} ⊆ d(G) Is a Dynkin system and G ⊆ D2 hence d(G) ⊆ D2 hence d(G) is a Dynkin system hence is a π-system S∞ • Let {A1 }∞ i=1 ∈ d(G), we need to show that i=1 Ai ∈ d(G) Si−1 define Bi = Ai \ j=1 Ai ∈ d(G) ∀i ∞ [ Ai = i=1 ∞ [ Bi ∈ d(G) i=1 since this is a disjoint union hence indeed d(G) is a σ-algebra Theorem: Let (S, F, µ) be a probability space and suppose F = σ(G) for some π-system G ∈ P(S) If ν is another probability measure on (S, F) s.t. µ(A) = ν(A) ∀A ∈ G then µ = ν Proof: write D = {A ∈ F : µ(A) = ν(A)} is a Dynkin system with G ⊆ D hence d(G) ⊆ D hence by the previous lemma F = σ(G) = d(G) = D Definition: Content µ is a content on (S, A) if • µ(φ) = 0 • µ(A ∪ B) = µ(A) + µ(B) ∀A, B ∈ A : A ∩ B = φ Theorem Caratheodory’s Theorem If µ is a σ-additive content on (S, A) where A is an algebra of S then µ extends to a measure on (S, σ(A)) Lemma: Cantor If {Kn }∞ n=1 are compact sets in a metric space s.t. Kn 6= φ ∀n and Kn+1 ⊆ Kn ∀n then ∞ \ Kn 6= φ n=1 Proof: Choose xn ∈ Kn for each n ∈ N since Kn 6= φ 3 1 {xn }∞ n=r Notice that ∈ Kr since Kn are decreasing ∃{xnk }∞ k=1 , x0 ∈ K1 s.t. lim xnk = x0 k→∞ by compactness {xnk }∞ k=r ∈ Kr T∞ hence x0 ∈ Kn ∀n hence x0 ∈ n=1 Kn Lemma: If {An }∞ n=1 ⊆ A and µ is a σ-additive content s.t. • A = a{Int(a, b) : −∞ ≤ a ≤ b ≤ ∞} where ( (a, b] b<∞ Int(a, b) = (a, ∞) b = ∞ • An+1 ⊆ An • limn→∞ An = φ • µ(An ) < ∞ ∀n then lim µ(An ) = 0 n→∞ Proof: By contradiction assume that limn→∞ µ(An ) = δ > 0 then µ(An ) ≥ δ ∀n ∈ N We can choose sets {Fn }∞ n=1 ∈ A s.t. • F n ⊆ An for each n • µ(An ) − µ(Fn ) ≤ 2−n δ • Fn+1 ⊆ Fn then we have that µ(Fn ) ≥ µ(An ) − δ2−n ≥ δ − δ2−n ≥ δ/2 hence Fn 6= φ moreover F n 6= φ F n is compact so by the previous lemma ∞ \ F n 6= φ n=1 but ∞ \ Fn ⊆ n=1 which is a contradiction to ∞ \ n=1 ∞ \ An 6= φ n=1 An = lim An = φ n→∞ Proposition: The Lebesgue measure L on (R, A) where A = a{Int(a, b) : −∞ ≤ a ≤ b ≤ ∞} extends to a measure on (R, σ(A) = (R, B(R)) Proof: By Caratheodory’s theorem it suffices to show that L is σ-additive 4 BASIC PROBABILITY 1 i.e. ∀{Ai }∞ i=1 ∈ A pairwise disjoint s.t. S∞ i=1 Ai ∈ A we have that ! ∞ ∞ [ X L Ai = L(Ai ) i=1 i=1 For Af inite := {A ∈ A : L(A) < ∞} S∞ suppose {Ai }∞ i=1 Sn ∈ Af inite be pairwise disjoint s.t. A = i=1 Ai ∈ Af inite Define Bn := i=1 Ai n X L(A) = L(A \ Bn ) + L(Ai ) i=1 by the previous lemma limn→∞ L(A \ Bn ) = 0 hence L ∞ [ ! Ai = i=1 ∞ X L(Ai ) i=1 so it remains to show this holds for A ∈ A \ Af inite suppose A = (−∞, a] S∞ let {Ai }∞ i=1 ∈ A be pairwise disjoint s.t. i=1 Ai = A Since {Ai }∞ are pairwise disjoint ∃ A ∈ {Ai }∞ 1 k i=1 i=1 s.t. Ak = (−∞, ak ] for some ak L(A) ≥ 0 since it is a content hence ∞ = L(Ak ) ≤ ∞ X L(Ai ) i=1 ∞ = L(Ak ) ≤ L ∞ [ ! Ai i=1 hence indeed the proposition holds. Definition: Measurable Function If (S, F), (S 0 , F 0 ) are measurable spaces then f : S → S 0 is (S, S 0 )-measurable if ∀A0 ∈ F 0 f −1 (A0 ) ∈ F Lemma: If (S, F), (S 0 , F 0 ) are measurable spaces and F 0 = σ(G), then it is sufficient that f −1 (A) ∈ F ∀A ∈ G for f to be measurable. Corollary: Let (S, F) is measurable and f : S → R then if {f ≤ a} ∈ F ∀a ∈ R then f is (S, R) measurable. Proof: By the previous lemma take G = {(−∞, a] : a ∈ R} which generates B(R) Lemma: If {fi }∞ i=1 are measurable then • h1 h2 • αh1 + βh2 • h1 ◦ h2 • infi hi • liminfi hi • limsupi hi 5 BASIC PROBABILITY 1 BASIC PROBABILITY are measurable when well defined. Corollary: Let S be a topological space with σ-algebra B(S) If f : S → R is continuous then it is Borel measurable. Proof: Open sets generate B(R) Since the preimage of open sets by a continuous function are open we have that ∀A open f −1 (A) is open hence is in the σ-algebra generated by open sets in S Definition: Random Variable If (Ω, F, P) is a probability space and (S, F 0 ) is measurable then X : Ω → S is a random variable if it is a measurable mapping i.e. ∀A0 ∈ F 0 X −1 (A0 ) ∈ F Definition: Push Forward Measure If X : Ω → S is a random variable on (Ω, F, P) then the distribution of X is the probability measure PX (A) = P({X ∈ A}) ∀A ∈ F 0 Definition: Cummulative Distribution Function Define FX (x) = PX ((−∞, x]) = P({X ≤ x}) Lemma: Suppose F = FX for some random variable X then • F : R → [0, 1] is non decreasing • F is right-continuous • limx→−∞ F (x) = 0 • limx→∞ F (x) = 1 Lemma: If F : R → [0, 1] s.t. • F : R → [0, 1] is non decreasing • F is right-continuous • limx→−∞ F (x) = 0 • limx→∞ F (x) = 1 then ∃(Ω, F, P) and random variable X s.t. FX = F Definition: Product σ-Algebra If (Si , Fi )ni=1 are measurable spaces then the product σ-algebra on these is n O Fi = σ({A1 × ... × An : Ai ∈ Fi }) i=1 Lemma: If (Ω, F), (Si , Fi )ni=1 are measurable spaces then {Xi : Ω → Si }ni=1 are all random variables N iff n Z := (X1 , ...Xn ) : Ω → S1 × ... × Sn is (F, i=1 Fi )-measurable Proof: Suppose Z is a random variable let πi : S1 × ... × Sn → Si be the ith canonical projection then Xi = πi ◦ Z 6 2 INDEPENDENCE Compositions of measurable functions are measurable so it is sufficient that the projections πi are measurable which is true since πi−1 (A) = R × ... × R × A × R × ... × R Nn Suppose {Xi }ni=1 are random variables and let A ∈ i=1 Fi Z −1 (A) = {ω : z(ω) ∈ A} = {ω : Xi (ω) ∈ Ai ∀i} = {ω : ω ∈ Xi−1 (Ai ) ∀i} n \ = Xi−1 Ai ∈ F i=1 Definition: Joint Distribution If {Xi }ni=1 : (Ω, F) → (Si , Fi ) are random variables for measurable spaces (Ω, F), (Si , Fi )ni=1 Nn then the distribution of Z = (X1 , ...Xn ) : (Ω, F) → (Sn × ... × Sn , i=1 Fi ) is called the joint distribution of (X1 , ...Xn ): n \ PZ (A) = P({Z ∈ A}) = P {Xi ∈ Ai } i=1 2 Independence Definition: Independence Let (Ω, F, P) be a probability space then • Sub-σ algebras {Fi }ni=1 are independent if whenever {ik }nk=1 are distinct and Aik ∈ Fik then n k \ Y Ai k = P(Aik ) P i=1 i=1 • Random variables {Xi }ni=1 are independent if the sub-σ algebras {σ(Xi )}ni=1 are independent where σ(Xi ) = σ({Xi−1 (A) : A ∈ Fi }) • Events {Ei }ni=1 ∈ F are independent if the sub-σ algebras {σ(Ei )}ni=1 are independent where σ(Ei ) = {φ, Ω, Ei , Ω \ Ei } Lemma: Suppose G, H are sub-σ algebras of F on the probability space (Ω, F, P) s.t. G = σ(I), H = σ(J ) for π-systems I, J then G, H are independent iff P(I ∩ J) = P(I)P(J) ∀I ∈ I, J ∈ J Proof: Independence =⇒ P(I ∩ J) = P(I)P(J) ∀I ∈ I, J ∈ J is trivial since independence =⇒ P(I ∩ J) = P(I)P(J) ∀I ∈ G ⊃ I, J ∈ H ⊃ J Define µ(A) = P(A ∩ I), ν(A) = P(A)P(I) ∀I ∈ I, A ∈ H clearly these are measures and by the property we have that they coincide on H Define η(A) = P(A ∩ H), κ(A) = P(A)P(H) ∀A ∈ G, H ∈ H clearly these are measures and by the property we have that they coincide on G 7 2 So indeed we have independence. Corollary: Let (Ω, F, P) be a probability space and X, Y : Ω → R be random variables. If P({X ≤ a} ∩ {Y ≤ b}) = P(X ≤ a)P(Y ≤ b) ∀a, b ∈ R then X, Y are independent. Proof: σ({(−∞, a] : a ∈ R}) = B(R) hence this holds by the previous lemma. Corollary: X, Y are independent iff FX (a)FY (b) = P(X ≤ a)P(Y ≤ b) = P({X ≤ a} ∩ {Y ≤ b}) = FX,Y (a, b) ∀a, b ∈ R Lemma: Let (Ωi , FN i , µi )i=1,2 be σ-finite measure spaces then ∃1 µ = µ1 µ2 which is σ-finite and µ(A1 × A2 ) = µ(A1 )µ2 (A2 ) ∀Ai ∈ Fi i = 1, 2 N µ is a measure on (Ω1 × Ω2 , F1 F2 ) Corollary: Random variables X, Y with joint distribution PX,Y are independent iff PX,Y = PX PY Definition: Tail-σ Algebras If {Fi }∞ i=1 are a sequence of σ-algebras then the tail-σ algebra is defined as T = ∞ ∞ [ \ {Fk } σ n=1 k=n If {Xi }∞ i=1 are a sequence of random variables then the tail-σ algebra is defined as T = ∞ ∞ [ \ σ {σ(Xk )} n=1 k=n Definition: Deterministic A random varibale X is deterministic if ∃c ∈ [−∞, ∞] s.t. P(X = c) = 1 Definition: P-Trivial T is P-trivial if • A∈T =⇒ P(A) ∈ {0, 1} • If X is a T -measurable random variable then X is deterministic. Lemma: Let µ be a measure on measurable space (Ω, F) and {Ai }∞ i=1 ∈ F • If Ai ⊆ Ai+1 ∀i then µ( lim Ai ) = lim µ(Ai ) i→∞ i→∞ • If Ai ⊇ Ai+1 ∀i and µ(A1 ) < ∞ then µ( lim Ai ) = lim µ(Ai ) i→∞ i→∞ 8 INDEPENDENCE 2 INDEPENDENCE Theorem: Kolmogorov Suppose that {Fi }∞ i=1 are an independent sequence of σ-algebras then the associated σ-algebra is P-trivial Proof: P(A) ∈ {0, 1} ⇐⇒ P(A)2 P(A) ⇐⇒ P(A)P(A) = P(A ∩ A) ⇐⇒ A ⊥ A S ∞ F Let Tn = σ k k=n+1 T∞ and T = n=1 Tn Sn Hn := σ k=1 Fk {Fk }∞ k=1 are independent hence Hn ⊥ Tn hence Hn ⊥ T therefore A ∈ T is independent from every Hn S ∞ n=1 Hn is a π-system hence it is sufficient that A⊥σ ∞ [ ∞ [ Hn = σ F k = T0 ⊃ T n=1 k=1 hence A ⊥ A Let c ∈ R by the first part of the theorem we have that P(X ≤ c) ∈ {0, 1} define c := sup{x : P(X ≤ x) = 0} If c = −∞ then P(X = −∞) = 1 If c = ∞ then P(X = ∞) = 1 If |c| < ∞ then 1 )=0 n 1 P(X ≤ c + ) = 1 n ∀n ∈ N P(X ≤ c − ∀n ∈ N hence by the previous lemma ∞ [ 1 = P(X < c) P {X ≤ c − n n=1 =0 ∞ \ 1 = P(X ≤ c) P {X ≤ c + n n=1 =1 P(X = c) = P(X ≤ c) − P(X < c) =1 Definition: Limsup Let (Ω, F, P) be a probability space with {Ai }∞ i=1 ∈ F then 9 2 {An : i.o.} = ∞ [ \ An m=1 n≥m = {ω : ∀m; ∃n ≥ m s.t. ω ∈ An } = limsupn An Definition: Liminf Let (Ω, F, P) be a probability space with {Ai }∞ i=1 ∈ F then {An : i.o.} = ∞ \ [ An m=1 n≥m = {ω : ∃m; ∀n ≥ m s.t. ω ∈ An } = liminfn An Lemma: Fatou Let (Ω, F, µ) be a measure space and {An }∞ n=1 ∈ F then • µ(liminfn An ) ≤ liminfn µ(An ) • If µ(Ω) < ∞ then µ(limsupn An ) ≥ limsupn µ(An ) Proof: • Define Zm := T n≥m An is monotonic and µ(Zm ) < µ(An ) ∀n ≥ m ∞ [ Zm µ(liminfn An ) = µ m=1 = lim µ(Zm ) m→∞ ≤ lim infn≥m µ(An ) m→∞ = liminfn µ(An ) • µ(Ω) − µ(limsupn An ) = µ(Ω \ limsupn An ) = µ(liminfn (Ω \ An )) ≤ liminfn µ(Ω \ An ) = µ(Ω) − limsupn µ(An ) Since µ(Ω) < ∞ we have that µ(limsupn An ) ≥ limsupn µ(An ) Lemma: Borel-Cantelli Let (Ω, F, P) be a probability space and {Ai }∞ i=1 ∈ F then 10 INDEPENDENCE 2 • If INDEPENDENCE P∞ P(An ) < ∞ then P(An : i.o.) = 0 P∞ • If {Ai }∞ i=1 are independent and n=1 P(An ) = ∞ then P(An : i.o.) = 1 n=1 Proof: S • define Gm : n≥m An T∞ then for k ∈ N we have m=1 Gm ⊂ Gk P({An : i.o.) = P(limsupn An ) ∞ \ = P( Gm ) m=1 ≤ P(Gk ) [ =P An ≤ ∞ X ∀k P(An ) n=k lim k→∞ ∞ X P(An ) = 0 n=k • {Ai }∞ i=1 independent =⇒ {Aci }∞ i=1 are independent. r r \ Y Acn = P(Acn ) P n=m = ≤ n=m r Y n=m r Y by independence 1 − P(An ) e−P(An ) n=m P − rn=m P(An ) =e lim e r→∞ − Pr n=m P(An ) =0 T ∞ c hence we have that P n=m An ) = 0 c (limsupn An ) = ∞ \ [ Acn m=1 n≥m is the countable union of null sets hence is itself a null set hence P(limsupn An ) = 1 Corollary: If {An }∞ n=1 are independent then P(An i.o.) ∈ {0, 1} Corollary: T∞ If we write Fn = σ(An ) then {An i.o.} is a tail event of T = n=1 σ({Fi }∞ i=n ) Lemma: Existence Of Independent Random Variables Let (Ωi , Fi , Pi ) i = 1, 2 be probability spaces with respective random variables X1 , X2 which have distributions PX1 , PX2 11 3 EXPECTATION AND CONVERGENCE N If we set Ω = Ω1 × Ω2 , F = F1 F2 , P = P1 × P2 then for ω = (ω1 , ω2 ) ∈ Ω define X̃1 (ω) = X1 (ω1 ) X̃2 (ω) = X2 (ω2 ) are defined on (Ω, F, P) and are independent. 3 Expectation And Convergence Definition: Simple Function A function f : S → R on measure space (S, F, µ) is simple if ∃{αi }ni=1 ∈ R|[0,∞] and partition {Ai }ni=1 ∈ F s.t. f= n X αi 1Ai i=1 Definition: Integration Of Simple Functions Pn If f : S → R is simple with f = i=1 αi 1Ai then Z f dµ = n X αi µ(Ai ) i=1 Definition: Integration Of Positive Measurable Functions If f : S → R|[0,∞] is measurable then Z o nZ gdµ : g ≤ f, g simple f dµ = sup Definition: µ-Integrable f : S → R is µ-integrable if Z |f |dµ < ∞ Definition: Integration Of µ-Integrable Functions If f : S → R is µ-integrable then Z Z Z f dµ = f + dµ − f − dµ Remark We denote the space of µ-integrable functions as L1 (µ) = L(S, F, µ) Lemma: The following properties hold for L1 (µ) • f, g ∈ L1 (µ), α, β ∈ R =⇒ Z Z αf + βgdµ = α Z f dµ + β • 0 ≤ f ∈ L1 (µ) =⇒ Z f dµ ≥ 0 • {fn }∞ n=1 , f measurable s.t. limn→∞ fn = f and fn ≤ fn+1 ∀n =⇒ Z Z lim fn dµ = f dµ n→∞ • f = 0 a.e. ⇐⇒ Z f dµ = 0 12 gdµ 3 EXPECTATION AND CONVERGENCE Theorem: Fubini’s N Theorem N N Let (S1 × S2 , F1N F2 , µ1 µ2 ) be a measure space and f ∈ L1 (µ1 µ2 ) then let µ = µ1 µ2 Z Z Z Z Z f (s1 , s2 )dµ1 (s1 )dµ2 (s2 ) = f (s)dµ(s) = f (s1 , s2 )dµ2 (s2 )dµ1 (s1 ) Moreover it is sufficient that f ≥ 0 measurable for the above integral inequality to hold. Definition: Expectation If X is a r.v. on (Ω, F, P) probability space X ∈ L1 (P) then Z E[X] := XdP Lemma: If X is a r.v. on (Ω, F, P) with distribution PX and h : R → R is Borel measurable then • h ◦ X ∈ L1 (P) ⇐⇒ h ∈ L1 (PX ) R R • E[h ◦ X] = h ◦ XdP = hdPX Corollary: Z E[X] = Z zdPX (z) = XdP Corollary: Take h = 1B for B ∈ B(R) then Z E[1B ◦ XdP] = 1B ◦ XdP Z = 1X −1 (B) dP = P(X −1 (B)) = P(X ∈ B) = PX (B) Z = 1B dPX This holds for all indicators hence by linearity holds for all simple functions therefore by monotone convergence theorem holds for all positive measurable functions and all integrable functions. Lemma: If X, Y are independent r.vs on probability space (Ω, F, P) then E[XY ] = E[X]E[Y ] Proof: Z E[XY ] = xydPX,Y Z Z = XY dPX dPY Z Z = Y XdPX dPY Z Z = XdPX Y dPY by Fubini = E[X]E[Y ] 13 3 EXPECTATION AND CONVERGENCE Remark: If X is a r.v on (Ω, F, P) and A ∈ F then we use the following notation: Z E[X; A] := E[X1A ] = X1A dP Lemma: Markov’s Inequality If X is a random variable, ε > 0 and g : R → R|[0,∞] is Borel measurable and non-decreasing then • P(g(X) > ε)ε ≤ E[g(X)] • P(X > ε)g(ε) ≤ E[g(X)] Proof: g ◦ X is measurable and non-negative by properties of measurable functions hence • Z E[g(X)] = g(X)dP Z ≥ g(X)1{g(X)>ε} dP Z ≥ ε1{g(X)>ε} dP Z ≥ ε 1{g(X)>ε} dP = εP(g(X) > ε) • Z E[g(X)] = g(X)dP Z ≥ g(X)1{X>ε} dP Z ≥ g(ε)1{X>ε} dP Z ≥ g(ε) 1{X>ε} dP = g(ε)P(X > ε) Corollary: Chebyshev’s Inequality If X is a r.v. with finite variance then P(|X − E[X]| > ε) ≤ ∀ε > 0 Proof: P(|X − E[X]| > ε) = P((X − E[X])2 > ε2 ) ≤ E[(X − E[X])2 ] ε2 V ar[X] = ε2 14 V ar[X] ε2 4 CONVERGANCE Theorem: If {Xi }ni=1 are independent random variables each with mean µ and variance σ 2 then for ε > 0 n X Xi σ2 P − µ > ε ≤ 2 n nε i=1 Proof: Pn define Z = i=1 Xni then E[Z] = µ V ar[Z] = σ 2 /n hence by Chebyshev’s inequality this clearly holds. Lemma: Jensen’s Inequality If X is a random variable with E[|X|] < ∞ s.t. f is convex on an interval containing the range of X and E[|f ◦ X|] < ∞ then f (E[X]) ≤ E[f (X)] Proof: Pn Suppose that X is simple so let X = i=1 ai 1Ai f (E[X]) = f n X ! ai P(Ai ) i=1 ≤ ∞ X P(Ai )f (ai ) by convexity i=1 = E[f (X)] Suppose that X is a positive r.v. then ∃{Xn }∞ n=1 r.vs which are increasing and converge to X a.s. then f (E[X]) = lim f (E[Xn ]) n→∞ ≤ lim E[f (Xn )] n→∞ = E[f (X)] 4 Convergance Definition: Almost Sure Convergance If {Xi }ni=1 are a sequence of r.vs on (Ω, F, P) then {Xi }ni=1 converge to r.v. X on (Ω, F, P) if P({ω ∈ Ω : lim Xn (ω) = X(ω)}) = 1 n→∞ Remark: Amost sure limits are unique only up to equality a.s. Corollary: X = Y a.s. iff P({ω ∈ Ω : X(ω) = Y (ω)}) = 1 Lemma: {Xn }∞ n=1 converge to X a.s. iff P(|Xn − X| > ε i.o.) = 0 ∀ε > 0, n ∈ N Corollary: Stability Properties • If {Xn }∞ n=1 are r.vs s.t. Xn → X a.s. and g : R → R ios continuous then g(Xn ) → g(X) a.s. 15 • If ∞ {Xn }∞ n=1 {Yn }n=1 4 CONVERGANCE are r.vs s.t. Xn → X a.s. and Yn → Y a.s. with α, β ∈ R then αXn + βYn → αX + βY Definition: Convergence In Probability If {Xn }∞ n=1 , X are r.vs on (Ω, F, P) then Xn → X in probability if lim P(|Xn − X| > ε) = 0 n→∞ Definition: Cauchy Convergence In Probability If {Xn }∞ n=1 , X are r.vs on (Ω, F, P) then Xn → X Cauchy in probability if lim P(|Xn − Xm | > ε) = 0 n,m→∞ Lemma: Let {Xn }∞ n=1 , X be r.vs on (Ω, F, P) If Xn → ∞ a.s. then Xn → X in probability Proof: Xn → ∞ a.s. implies ∀ε > 0 P(|Xn − X| > ε) ≤ P(supk≥n |Xk − X| > ε) hence we have lim P(|Xn − X| > ε) ≤ lim P(supk≥n |Xk − X| > ε) n→∞ n→∞ ≤ P(limn→∞ supk≥n |Xk − X| > ε) = P(|Xn − X| > ε i.o.) =0 Lemma: Let {Xn }∞ n=1 , X be r.vs on (Ω, F, P) Xn → X in probability iff Xn → X Cauchy in probability. Proof: Suppose Xn → X in probability. then |Xn − Xm | ≤ |Xn − X| + |X − Xm | {|Xn − Xm | > ε} ⊆ {|Xn − X| > ε/2} ∪ {|X − Xm | > ε/2} P(|Xn − Xm | > ε) ≤ P(|Xn − X| > ε/2) + P(|X − Xm | > ε/2) lim P(|Xn − Xm | > ε) ≤ n,m→∞ lim P(|Xn − X| > ε/2) + P(|X − Xm | > ε/2) n,m→∞ =0 Now suppose Xn → X Cauchy in probability for any k ≥ 1 choose ε = 2−k ∃mk s.t. ∀n > m ≥ mk we have that P(|Xn − Xm | > 2−k ) < 2−k set n1 = m1 and recursively define nk+1 = max{nk + 1, mk+1 } define Xk0 := Xnk 0 definePAk := {|Xk+1 − Xk0 | > 2−k } ∞ then k=1 P(Ak ) < ∞ then by Borel Cantelli 0 |Xk+1 (ω) − Xk0 (ω)| ≤ 2−k ∀ω ∈ Ω \ A, where P(A) = 0 hence ∀ω ∈ Ω \ A 16 ∀k ≥ K0 4 lim n→∞ 0 supm>n |Xm − Xn0 | ≤ lim n→∞ ∞ X 0 |Xm − Xn0 | k=n ∞ X 2−k ≤ lim n→∞ k=n =0 hence P(limsupn Xn0 = liminfn Xn0 < ∞) = 1 define X := liminfn Xn0 since Xnk → X a.s. by the previous lemma we have that Xnk → X in probability hence ∀ε > 0 lim P(|Xk − X| > ε) ≤ lim (P(|Xk − Xnk | > ε/2) + P(|Xnk − X| > ε/2)) k→∞ k→∞ =0 Lemma: LetP{Xn }∞ n=1 , X be random variables on (Ω, F, P) ∞ If n=1 P(|Xn − X| > ε) < ∞ ∀ε > 0 then Xn → X a.s. Proof: This follows directly from Borel Cantelli. Lemma: If {Xn }∞ n=1 , X are r.vs then Xn → X in probability iff ∞ ∀{Xnk }∞ k=1 ∃{Xnkr }r=1 s.t. Xnkr → X a.s. Proof: Suppose Xn → X in probability the ∀ε > 0 find nkr ∈ N s.t. P(|Xnkr − X| > 1/r) < 2−r then ∞ ∞ X X P(|Xnkr − X| > 1/r) < 2−r < ∞ r=1 r=1 moreover ∀ε > 0 X r>1/ε P(|Xnkr − X| > ε) < X P(|Xnkr − X| > 1/r) r>1/ε ≤ X 2−k r>1/ε <∞ hence we simply take the sequence {Xnkr }r>1/ε which converges to X a.s. Assume that every subsequence has a subsequence converging a.s. we need to show that X is unique. fix ε > 0 bn := P(|Xn − X| > ε) ∈ [0, 1] is compact hence 17 CONVERGANCE 4 ∃bnk converging subsequence by our assumption Xnk has a subsequence Xnkr → X a.s. lim bnkr = lim P(|Xnkr − X| > ε) = 0 r→∞ r→∞ since a.s. convergence implies convergence in probability. hence since bnk is convergent we must have that limk→∞ bnk = 0 hence every subsequence of bn which converges does so to 0 Theorem: The Weak Law Of Large Numbers 1 If {Xn }∞ n=1 are i.i.d r.vs with Xi ∈ L (Ω, F, P) then lim n→∞ n X Xk k=1 n = E[X1 ] Proof: Write Xk∗ := Xk 1{|Xk |≤k1/4 } Xk0 := Xk 1{|Xk |>k1/4 } Yn∗ := Yn0 := n X X ∗ − E[X ∗ ] k k=1 n X k=1 k n Xk0 − E[Xk0 ] n It suffices to show that Yn∗ , Yn0 → 0 in probability. V ar[Xk∗ ] ≤ E[Xk∗ ]2 V ar[Yk∗ ] ≤ ≤ k 1/2 n X V ar[X ∗ ] k n2 k=1 ≤ n X k 1/2 k=1 ≤ n2 n X n1/2 k=1 n2 = n−1/2 by Chebyshev’s inequality P(|Yn∗ | > ε) ≤ but V ar[Yn∗ ] ε2 V ar[Yn∗ ] =0 n→∞ ε2 lim hence Yn∗ → 0 in probability E[|Xk0 |] = E[|X10 |1|X10 |>k1/4 ] → 0 by MCT 18 CONVERGANCE 4 CONVERGANCE hence E[|Yn0 |] εP P(|Yn0 | ≥ ε) ≤ ≤ 2 n n k=1 E[|Xk0 |] ε lim P(|Yn0 | ≥ ε) = 0 n→∞ hence Yn0 → 0 in probability. Theorem: The Strong Law Of Large Numbers let {Xn }∞ n=1 be independent r.vs s.t. • E[|Xn |2 ] < ∞ ∀n • v := supXn {V ar[Xn ]} < ∞ then n X Xk − E[Xk ] n k=1 →0 a.s. Proof: WLOG assume Pn E[Xk ] = 0 define Sn = i=1 Xni Notice that Sn → 0 in probability so we want a subsequence Snk → 0 a.s. P(|Sn2 | > ε) ≤ hence ∞ X V ar[Sn2 ] n2 ε2 P(|Sn2 | > ε) < ∞ ∀ε > 0 n=1 so we have that Sn2 → 0 a.s. For m ∈ N let n = nm s.t. n2 ≤ m ≤ (n + 1)2 Pk write yk = ksk = i=1 Xi P(|ym − yn2 | > εn2 ) ≤ ε−2 n−4 V ar m h X Xi i i=n2 +1 2 ≤ ε−2 n−4 v(m − n ) 2 ∞ (n+1) −1 v X X m − n2 P(|ym − yn2m | > εn ) ≤ 2 ε n=1 n4 2 m=1 ∞ X 2 m=n 2n X k v X n = 1∞ ≤ 2 ε n4 k=1 ∞ v X 2n(2n + 1) = 2 ε n=1 2n4 ≤ ∞ v X 2 ε2 n=1 n2 <∞ hence since the intersection of two sets of full measure is also a set of full measure (where finite) 19 4 we get that |Sm | = |ym /m| ≤ |ym /n2 | → 0 a.s. Definition: Absolute Moment Let X be a random variable on probability space (Ω, F, P) then for p ∈ [1, ∞) the pth moment of X is defined as Z p E[|X| ] = |X|p dP Definition: Lp Space Lp (Ω, F, P) := {X : Ω → R : E[|X|p ] < ∞, X measurable} Definition: Lp Norm The Lp norm is defined as ||X||p := E[|X|p ]1/p Lemma: Holder’s Inequality For p, q ∈ [1, ∞) s.t. p1 + 1q = 1 we have that for X ∈ Lp , Y ∈ Lq E[|XY |] ≤ ||X||p ||Y ||q moreover this holds for p = 1, q = ∞ where ||Y ||∞ := inf {s ≥ 0 : P(|Y | > s) = 0} Lemma: For p ∈ [1, ∞], X, Y ∈ Lp we have ||X + Y ||p ≤ ||X||p + ||Y ||p Corollary: ||X||p = 0 =⇒ X = 0 a.s. Definition: Quotient Space Define N0 = {X ∈ Lp : X = 0 P a.s.} then the quotient space of Lp is Lp = Lp N0 Definition: Equivalence Class We denote the equivalence class of X ∈ Lp to be [X] := {Y : X = Y a.s.} Definition: Convergance in Lp p Let p ∈ [1, ∞] and {Xn }∞ n=1 , X ∈ L then Xn → X in Lp if lim ||Xn − X||p = 0 n→∞ Definition: Cauchy Sequence p {Xn }∞ n=1 is a Cauchy sequence in L if ∀ε > 0 ∃N ∈ N s.t. ∀n, m ≥ N we have that ||Xn − Xm ||p < ε Theorem: 20 CONVERGANCE 4 {Xn }∞ n=1 p CONVERGANCE p for p ∈ [1, ∞] let ∈ L be a Cauchy sequence in L then ∃X ∈ Lp s.t. Xn → X in Lp Lemma: If 1 ≤ s ≤ r then Xn → X in Lr =⇒ Xn → X in Ls Proof: For s = 1, r = 2 E[|Xn − X|s ] = Z |Xn − X|dP Z |Xn − X|dP = Z ≤ 1/2 Z 1/2 |Xn − X|2 dP 12 dP by Cauchy-Schwarz inequality = E[|Xn − X|2 ]1/2 lim E[|Xn − X|2 ] = 0 =⇒ n→∞ lim E[|Xn − X|2 ]1/2 = 0 n→∞ hence indeed the lemma holds for s = 1, r = 2 Lemma: p Let {Xn }∞ n=1 , X be r.vs s.t. Xn → X in L for p ≥ 1 then Xn → X in probability. Proof: let ε > 0 P(|Xn − X| > ε) = P(|Xn − X|p > εp ) ≤ E[|Xn − X|p ] εp by Markov’s inequality p lim n→∞ E[|Xn − X| ] =0 εp by convergence in Lp Lemma: X ∈ L1 (Ω, F, P) iff Z |X|dP = 0 lim k→∞ {|X|>k} Definition: Uniformly Integrable A set of random variables C on (Ω, F, P) is uniformly integrable if Z lim supX∈C |X|dP = 0 k→∞ {|X|>k} Theorem: 1 If X, {Xn }∞ n=1 ∈ L (Ω, F, P) then 1 Xn → X in L iff • Xn → X in probability • {Xn }∞ n=1 is uniformly integrable. 21 4 CONVERGANCE Proof: Xn → X in L1 =⇒ Xn → X in probability by a previous lemma. moreover if Xn → X in L1 we have finite integrability since otherwise P(X = ∞) > 0 hence the expectation cannot be finite. Assume that Xn → X in probability and {Xn }∞ n=1 is uniformly integrable. for k > 0 define x>k k ϕk (x) = x x ∈ [−k, k] −k x < −k then we have that E[|Xn − X|] ≤ E[|Xn − ϕk (Xn )|] + E[|ϕk (Xn ) − ϕk (X)|] + E[|ϕk (X) − X|] notice that |ϕk (Xn ) − Xn | = 1{|Xn |>k} |k − |Xn | ≤ 2|Xn |1{|Xn |>k} hence let ε > 0 then by uniform integrability we have that ∃k1 ∈ [0, ∞) s.t. ∀k ≥ k1 Z supn |Xn |dP ≤ ε/6 {|Xn |>k} hence ∀n we have that k > k1 =⇒ Z E[|Xn − ϕk (Xn )|] ≤ supn 2 |Xn |dP ≤ ε/3 {|Xn |>k} Since X ∈ L1 we have that Z |X|dP = 0 lim k→∞ {|X|>k} hence ∃k2 ∈ [0, ∞) s.t. ∀k > k2 E[|X − ϕk (X)|] ≤ ε/3 set k0 = min{k1 , k2 } P(ϕk0 (Xn ) − ϕk0 (X) > ε) ≤ P(Xn − X > ε) hence since Xn → X in probability we have that ϕk (Xn ) → ϕk (X) in probability so by the Dominated Convergence Theorem |ϕk0 | ≤ k0 so ∃N ∈ N s.t. ∀n ≥ N we have that E[|ϕk (Xn ) − ϕ(X)|] < ε/3 hence indeed we have that for k = k0 and n ≥ N we have that E[|Xn − X|] ≤ E[|Xn − ϕk (Xn )|] + E[|ϕk (Xn ) − ϕk (X)|] + E[|ϕk (X) − X|] ≤ ε Definition: Probability Set We define Prob(R) to be the set of probability measures on (R, B(R)) Definition: Continuous Bounded Functions We define Cb (R) to be the set of continuous bounded functions f : R → R Definition: Weak Convergence • If µ, {µn }∞ n=1 ∈ Prob(R) then µn → µ weakly if Z lim n→∞ Z f dµn = f dµ • If F, {Fn }∞ n=1 are distribution functions then Fn → F weakly if µn → µ weakly where µ, {µn }∞ n=1 are the associated probability measures F (x) = µ((−∞, x]) 22 ∀f ∈ Cb (R) 4 CONVERGANCE X, {Xn }∞ n=1 • If are random variables then Xn → X weakly if PXn → PX weakly where PX (A) = P(X ∈ A) Definition: Convergence In Distribution • If F, {Fn }∞ n=1 are distribution functions then Fn → F in distribution if for all continuous points x of F lim Fn (x) = F (x) n→∞ • If X, {Xn }∞ n=1 are random variables then Xn → X in distribution if FPXn → FPX in distribution where FPX (x) = P(X ≤ x) Theorem: ∞ If {Fn }∞ n=1 , F are distribution functions of probability measures {µn }n=1 µ then Fn → F weakly iff Fn → F in distribution. Proof: Suppose Fn → F weakly fix x ∈ R s.t. F is continuous at x and let δ > 0 define y≤x 1 y−x h(y) = 1 − δ y ∈ (x, x + δ) 0 y ≥x+δ and then since Fn (x) = 1 g(y) = 1 − 0 R y ≤x−δ y ∈ (x − δ, x) y≥x y−x δ 1(−∞,x] dµn we have that Z Z gdµn ≤ F (x) ≤ hdµn Z limsupn Fn (x) ≤ limsupn Z Z liminfn Fn (x) ≥ liminfn hdµ ≤ F (x + δ) hdµn = Z gdµn = gdµ ≥ F (x − δ) by continuity of h, g By continuity of F at x we have that lim F (x + δ) = F (x) = lim F (x − δ) δ→0 δ→0 hence lim Fn (x) = F (x) n→∞ hence Fn → F in distribution. Suppose Fn → F in distribution. then Z lim n→∞ Z 1(−∞,x] dµn = 1(−∞,x] dµ ∀x ∈ R Since any continuous bounded function can be approximated by sums of indicator functions and linearity of the integral we indeed have that Fn → F weakly Lemma: Let {Xi }∞ i=1 , X be random variables on (Ω, F, P) then if Xn → X in probability then Xn → X weakly Proof: There are two different proofs of this 23 4 CONVERGANCE • Since Xn → X in probability and f is continuous we have that f (Xn ) → f (X) in probability. f (Xn ) is bounded by the proerties of f hence by the dominated convergance theorem Z Z f (Xn )dP = f (X)dP lim n→∞ hence indeed Xn → X weakly. • Since f is bounded we know that ∃k > 0 s.t. |f (Xn )| ≤ k Z limk→∞ supXn ∀n hence |f (Xn )|dP = 0 {|f (Xn )>k|} so f (Xn ) is uniformly integrable. Since f is continuous and Xn → X in probability we have that f (Xn ) → f (X) in probability. This along with uniform integrability implies that f (Xn ) → f (X) in L1 i.e. limn→∞ E[|f (Xn ) − f (X)|] = 0 hence indeed we have weak convergance. Corollary: Let {Xi }∞ i=1 , X be random variables on (Ω, F, P) then if Xn → X a.s. then Xn → X weakly Definition: Conditional Probability If (Ω, F, P) is a probability space with A, B ∈ F then the conditional probabiltiy of B given A has occurred (where P(A) > 0 is P(B|A) = P(A ∩ B) P(A) Definition: Conditional Expectation If (Ω, F, P) is a probability space with A ∈ F with strictly positive probability then the conditional expectation of r.v X given A has occurred is Z 1 E[X|A] = XdP P(A) A morover if F0 = σ(G) for a countable partition G = {Ai }∞ i=1 of measurable sets on Ω then Z X 1Ai E[X|F0 ] = XdP P(Ai ) Ai i:P(Ai )>0 is a random variable. Theorem: Let X ∈ L1 (Ω, F, P), F0 = σ(G) s.t. G is a countable partition {Ai }∞ i=1 ∈ Ω Then E[X|F0 ] has the following properties: • E[X|F0 ] is F0 measurable. • ∀A ∈ F0 we have that E[X1A ] = E[1A E[X|F0 ]] Proof: i] • This is trivial since E[X;A P(Ai ) is constant, 1Ai are measurable since Ai ∈ F0 and the countable sum of measurable functions is measurable. • Let A ∈ G with P(A) > 0 24 4 CONVERGANCE 1 A i E[1A E[X|F0 ]] = E 1A E[X; Ai ] P(Ai ) i:P(Ai )>0 1A E[X; A] =E P(A) E[1A ]E[X; A] = P(Ai ) = E[X; A] X since {Ai }∞ i=1 form a partition = E[X1A ] This extends to A ∈ F0 by standard operations of elements of σ-algebras. If P(A) = 0 then X1A , E[X|F0 ]1A = 0 a.s. hence both sides of the equation are null. Corollary: Taking y = 1 in the second property we have that E[X] = E[E[X|F0 ]] Definition: Version Of Conditional Expectation A random variable X0 ∈ L1 (Ω, F, P) is called a version E[X|F0 ] if • X0 is F0 measurable • ∀A ∈ F0 we have E[1A X] = E[1A X0 ] Theorem: Let X1 , X2 ∈ L1 (Ω, F, P) then • E[X1 + X2 |F0 ] = E[X1 |F0 ] + E[X2 |F0 ] P a.s. • ∀c ∈ R E[cX|F0 ] = cE[X|F0 ] P a.s. Proof: For any choice of E[Xi |F0 ] we have that E[X1 |F0 ] + E[X2 |F0 ] is measurable hence E[1A (E[X1 |F0 ] + E[X2 |F0 ])] = E[1A E[X1 |F0 ] + 1A E[X2 |F0 ]] = E[1A E[X1 |F0 ]] + E[1A E[X2 |F0 ]] = E[1A X1 ] + E[1A X2 ] = E[1A (X1 + X2 )] = E[1A E[X1 + X2 |F0 ]] Theorem: Let X1 , X2 ∈ L1 (Ω, F, P) s.t. X1 ≤ X2 then E[X1 |F0 ] ≤ E[X2 |F0 ] P a.s. 25 by linearity of expectation 4 Proof: Let B0 := {ω ∈ Ω : E[X1 |F0 ] > E[X2 |F0 ]} ∈ F0 we want to show that P(B0 ) = 0 0 ≤ E[1B0 (E[X1 |F0 ] − E[X2 |F0 ])] = E[1B0 (X1 − X2 )] ≤0 hence P(B0 ) = 0 as required. Theorem: If Z, W are conditional expectations of X|F0 for some X ∈ L1 then Z = W Proof: Z, W are F0 -measurable so write A0 := {Z > W } ∈ F0 then we have P a.s. E[1A0 (Z − W )] = E[1A0 Z] − E[1A0 W ] = E[1A0 X] − E[1A0 X] =0 since 1A0 (Z − W ) ≥ 0 we have that 1A0 (Z − W ) = 0 a.s. but since Z > W on A0 it must be that 1A0 = 0 a.s. hence P(A0 ) = 0 this holds similarly for A1 := {W > Z} P(Z 6= W ) = P({Z > W } ∪ {Z < W }) ≤ P(A0 ) + P(A1 ) =0 Definition: Respective Density If µ, ν are measures on measure space (Ω, F) then ν has density with respect to µ if ∃f : Ω → [0, ∞) that is F-measurable s.t. ∀A ∈ F we have Z ν(A) = f dµ A Theorem: Radon-Nikodym If µ, ν are finite measures on measurable space (Ω, F) then the following are equivalent: • µ(A) = 0 =⇒ ν(A) = 0 • ν has density w.r.t. µ Lemma: If 0 ≤ X ∈ L1 and F0 ⊆ F is a sub-σ-algebra then take µ := P|F0 and ν : F0 → [0, ∞) s.t. Z ∀A ∈ F0 ν(A) = XdP A then if A is µ-null then by Radon-Nikodym ν has density g w.r.t. µ Lemma: 26 CONVERGANCE 4 CONVERGANCE If g is the density of ν w.r.t. µ where Z ν(A) := XdP A then g is a conditional expectation E[X|F0 ] Proof: let A0 ∈ F0 Z E[g1A0 ] = gdP ZA0 = gdµ A0 Z = 1A0 dν Z = 1A0 XdP = E[1A0 X] Corollary: If X ∈ L1 then X = X+ − X− and if g+ , g− are versions of E[X+ |F0 ], E[X− |F0 ] then by linearity E[X|F0 ] = g+ − g− Theorem: If X ∈ L1 (Ω, F, P) and F0 ⊆ F1 ⊆ F are sub-σ-algebras then X0 := E[E[X|F1 ]|F0 ] = E[X|F0 ] Proof: We want to show that E[X0 1A0 ] = E[X1A0 ] E[X0 1A0 ] = E[1A0 E[E[X|F1 ]|F0 ]] = E[1A0 E[X|F1 ] since A0 ∈ F0 = E[X1A0 ] since A0 ∈ F1 Theorem: If X ∈ L1 (Ω, F, P) then E[X|F0 ] ∈ L1 (Ω, F, P) Proof: Define A0 := {E[X|F0 ] ≥ 0} ∈ F0 then E[E[X|F0 ]+ ] = E[E[X|F0 ]1A0 ] = E[X1A0 ] since X ∈ L1 <∞ this holds similarly for E[E[X|F0 ]− ] hence this holds for E[X|F0 ] by linearity. Theorem: 27 4 CONVERGANCE Let (Ω, F, P) be a probability space with F0 ⊂ F sub-σ algebra, X ∈ L (Ω, F, P) where F0 ⊥ σ(X) then E[X|F0 ] = E[X] P a.s. 1 Proof: E[X] is F0 measurable since constant so let A ∈ F0 E[1A E[X]] = E[X]E[1A ] since E[X] is constant = E[1A X] by independence = E[1A E[X|F0 ]] Theorem: MCT For Conditional Expectations 1 Let {Xn }∞ n=1 ∈ L be an increasing sequence s.t. Xn → X then P-a.s. for some X ∈ L1 E[Xn |F0 ] → E[X|F] both P-a.s. and in L1 Proof: Since E[Xn |F0 ] is an increasing sequence of random variables ∃Z which is an F0 measurable r.v. s.t. E[Xn |F0 ] → Z P-a.s. For A ∈ F0 we have that E[Z1A ] = E[ lim E[Xn |F0 ]1A ] n→∞ = lim E[E[Xn |F0 ]1A ] by Monotone Convergance of E n→∞ = lim E[Xn 1A ] n→∞ = E[X1A ] by MCT hence Z = E[X|F0 ] P-a.s. 0 ≤ E[X|F0 ] − E[Xn |F0 ] ≤ E[X|F0 ] − E[X1 |F0 ] P a.s. 1 ∈L hence by dominated convergence theorem lim E[|E[X|F0 ] − E[Xn |F0 ]|] = 0 n→∞ Theorem: DCT For Conditional Expectations 1 Let {Xn }∞ P-a.s. n=1 ∈ L converge to X 1 If ∃Z ∈ L s.t. ∀n ∈ N we have that |Xn | ≤ Z P-a.s. then • X ∈ L1 • E[Xn |F0 ] → E[X|F0 ] P-a.s. • E[Xn |F0 ] → E[X|F0 ] ∈ L1 Proof: • X ∈ L1 since |Xn | ≤ Z P-a.s. =⇒ |X| ≤ Z P-a.s. 28 4 CONVERGANCE • define Un := infm≥n Xm is an increasing sequence and Vn := supm≥n Xm is a decreasing sequence then Un , Vn → X P-a.s. moreover −Z ≤ Un ≤ Xn ≤ Vn ≤ Z by MCT for conditional expectations we have that E[Un |F0 ] → E[X|F0 ] E[Vn |F0 ] → E[X|F0 ] both P-a.s. and in L1 and by monotonicity of conditional expectations we have that E[Un |F0 ] ≤ E[Xn |F0 ] ≤ E[Vn |F0 ] so indeed E[Xn |F0 ] → E[X|F0 ] in L1 and P-a.s. Theorem: Let X ∈ L1 (Ω, F, P) and Y be F0 measurable s.t. XY ∈ L1 then E[XY |F0 ] = Y E[X|F0 ] Proof: Suppose Y = 1A for some A ∈ F0 then take B ∈ F0 E[1B Y E[X|F0 ]] = E[1B∩A E[X|F0 ]] since A ∩ B ∈ F0 = E[1B∩A X] = E[1B Y X] = E[1B E[XY |F0 ]] hence the theorem P∞ holds for indicator functions Suppose Y = n=1 αn 1An " E[XY |F0 ] = E ∞ X # αn 1An X|F0 n=1 = = ∞ X n=1 ∞ X αn E[1An X|F0 ] by linearity αn 1An E[X|F0 ] by the first part of the theorem n=1 = Y E[X|F0 ] hence the theorem holds for simple functions Suppose Y is positive let {Yn }∞ n=1 be an increasing sequence of simple functions s.t. Yn → Y |Yn X| ≤ |Y X| ∈ L1 and Yn X → Y X hence by DCT for conditional expectations we have that 29 4 Yn E[X|F0 ] = E[Yn X|F0 ] → E[Y X|F0 ] CONVERGANCE P − a.s. moreover Yn E[X|F0 ] → Y E[X|F0 ] P − a.s. hence the theorem holds for positive functions. For the general case write Y = Y+ − Y− E[XY |F0 ] = E[X(Y+ − Y− )|F0 ] = E[XY+ |F0 ] − E[XY− |F0 ] = Y+ E[X|F0 ] − Y− E[X|F0 ] = Y E[X|F0 ] Lemma: If X ∈ L1 , F0 ⊆ F is a sub-σ algebra then ||E[X|F0 ]||1 ≤ ||X||1 Proof: Z ||E[X|F0 ]||1 = |E[X|F0 ]|dP Z = Z = Z E[X|F0 ]+ dP + E[X|F0 ]− dP Z E[X|F0 ]1E[X|F0 ]>0 dP + E[X|F0 ]1E[X|F0 ]<0 dP = E[X1E[X|F0 ]>0 ] + E[X1E[X|F0 ]<0 ] ≤ E[|X|] Z = |X|dP = ||X||1 Theorem: Conditional Jensen’s Inequality Let I ⊂ R be open, ϕ : I → R be convex and X ∈ L1 a r.v. with X : Ω → I hence ϕ ◦ X ∈ L1 then E[ϕ ◦ X|F0 ] ≥ ϕ(E[X|F0 ]) Proof: From Jensen’s inequality we have that ϕ(X) ≥ ϕ(E[X|F0 ]) + D− ϕ(E[X|F0 ])(X − E[X|F0 ]) let An = {ω ∈ Ω : |D− ϕ(E[X|F0 ])| ≤ n} ∈ F0 then since 1An − F0 measurable 1An E[ϕ(X)|F0 ] = E[1An ϕ(X)|F0 ] ≥ E[1An ϕ(E[X|F0 ])] + E[1An D− ϕ(E[X|F0 ])(X − E[X|F0 ])] = 1An ϕ(E[X|F0 ]) + 1An D− ϕ(E[X|F0 ])E[X − E[X|F0 ]] = 1An ϕ(E[X|F0 ]) + 1An D− ϕ(E[X|F0 ])(E[X|F0 ] − E[X|F0 ] = 1An ϕ(E[X|F0 ]) 30 4 CONVERGANCE this holds for all n hence also for limn→∞ An = Ω Theorem: Let p ≥ 1, X ∈ Lp then ||E[X|F0 ]||p ≤ ||X||p Proof: Write ϕ(x) = xp Z ||E[X|F0 ]||p = Z = Z ≤ is convex hence by Jensen’s inequality |E[X|F0 ]|p dP ϕ(|E[X|F0 ]|)dP |E[ϕ(X)|F0 ]|dP Z = |E[X p |F0 ]|dP = ||X||p Corollary: If (Ω, F), (Ω0 , F 0 ) are measurable spaces, Y : (Ω, F) → (Ω0 , F 0 ) is F − F 0 measurable and Z : Ω → R then Z is σ(Y ) measurable iff ∃g : (Ω0 , F 0 ) → (R, B(R)) s.t. Z = g ◦ Y Proof: Clearly if ∃g : (Ω0 , F 0 ) → (R, B(R)) s.t. Z = g ◦ Y then Z is σ(Y ) measurable since the composition of measurable functions is measurable. Suppose n X Z= αk 1Ak k=1 {Ak }nk=1 for ∈ σ(Y ) notice that σ(Y ) = {Y −1 (A0k ) : A0k ∈ F 0 } hence Ak = Y −1 (A0k ) for some A0k ∈ F 0 so Z= = = = n X k=1 n X k=1 n X αk 1Ak αk 1Y −1 (A0k ) αk (1A0k ◦ Y ) k=1 n X ! αk 1A0k ◦Y k=1 =g◦Y where clearly g is F 0 − B(R) measurable since {A0k }nk=1 ∈ F 0 Suppose Z is positive. then ∃{Zn }∞ n=1 simple increasing functions s.t. limn→∞ Zn = Z 0 0 then ∃{gn }∞ n=1 (Ω , F ) → (R, B(R)) s.t. Zn = gn ◦ Y then define g = supn gn is clearly F 0 − B(R) measurable and Z = g ◦ Y 31 4 For general Z write Z = Z+ − Z− then ∃g+ , g− (Ω0 , F 0 ) → (R, B(R)) s.t. Z+ = g+ ◦ Y, Z− = g− ◦ Y so write g = g+ − g− hence Z = (g+ − g− ) ◦ Y = g ◦ Y Definition: Factorised Conditional Expectation f : (R, B(R)) → (R, B(R)) is a FCE of the r.v. X given the r.v. Y : Ω → R if f ◦ Y is a version of the conditional expectation of X given Y i.e. E[X|Y ] = E[X|σ(Y )] = f (Y ) Corollary: Let Y : (Ω, F) → (R, B(R)) and X ∈ L1 (Ω, F, P) then ∃f FCE of X given Y and Z f dPY = E[1Y −1 (A) X] A Proof: By the previous lemma write Z = E[X|Y ] which is σ(Y ) measurable hence indeed ∃f FCE of X given Y moreover Z Z f dPY = 1Y −1 (A) XdP A = E[1Y −1 (A) X] Remark: E[X|Y = y] is unique PY -a.s. Theorem: Let Y : (Ω, F) → (R, B(R)) and X ∈ L1 (Ω, F, P) If λ, µ are σ-finite measures on (R, B(R)) s.t. (X, Y ) has density g w.r.t. λ ⊗ µ then R xg(x, y)dλ(x) f (Y ) := RR g(x, y)dλ(x) R is an FCE of X given Y Proof: recall that Z P((X, Y ) ∈ A1 × A2 ) = g(x, y)d(λ ⊗ µ)(x, y) ZA1 ×A2 E[h(X, Y )] = h(x, y)g(x, y)d(λ ⊗ µ)(x, y) R2 Suppose f takes the required form, hence clearly it is Borel measurable. 32 CONVERGANCE 5 MARTINGALES Let A ∈ B(R) Z x1A (y)g(x, y)d(λ ⊗ µ)(x, y) Z Z = 1A (y) xg(x, y)dλ(x)dµ(y) Z Z = 1A (y)f (y) g(x, y)dλ(x)dµ(y) Z = 1A (y)f (y)g(x, y)d(λ ⊗ µ)(x, y) E[X1Y −1 (A) ] = by Fubini = E[1A (y)f (y)] hence indeed f (y) = E[X|Y ] 5 Martingales Definition: Filtration If (Ω, F) is a measurable space then {Fi }∞ i=1 sub-σ-algebras of F are a filtration if Fi ⊆ Fi+1 ⊆ F Definition: Adapted ∞ A sequence of random variables {Xi }∞ i=1 is called adapted to {Fi }i=1 if Xn is Fn -measurable ∀n Definition: Martingale If (Ω, F, P) is a probability space and {Fi }∞ i=1 is a filtration of (Ω, F) then {Xn }∞ is a martingale if: n=1 • E[|Xn |] < ∞ ∀n ∞ • {Xn }∞ n=1 are adapted to {Fn }n=1 • E[Xn+1 |Fn ] = Xn Remark: If {Xn }∞ n=1 has the first three properties of a martingale but E[Xn+1 |Fn ] ≥ Xn then {Xn }∞ n=1 is a sub-martingale E[Xn+1 |Fn ] ≤ Xn then {Xn }∞ n=1 is a super-martingale Lemma: ∞ If {Xn }∞ n=1 is a martingale w.r.t. {Fn }n=1 then E[Xn+m |Fn ] = 0 Proof: E[Xn+m |Fn ] = E[E[Xn+m |Fn ∩ Fn+m−1 ]] = E[E[Xn+m |Fn+m−1 ]|Fn ] = E[Xn+m−1 |Fn ] = E[Xn+1 |Fn ] = Xn Theorem: 2 ∞ If {Xn }∞ n=1 ∈ L is a martingale w.r.t. {Fn }n=1 and r ≤ s ≤ t then • E[Xr (Xs − Xr )|Fr ] = 0 • E[Xr (Xt − Xs )|Fs ] = 0 33 ∀i 5 • E[(Xt − Xs )(Xs − Xr )|Fs ] = 0 • E[Xr (Xs − Xr )] = 0 • E[Xr (Xt − Xs )] = 0 • E[(Xt − Xs )(Xs − Xr )] = 0 Proof: • E[Xr (Xs − Xr )|Fr ] = E[Xr Xs − Xr2 |Fr ] = E[Xr Xs |Fr ] − E[Xr2 |Fr ] = X r E[Xs |Fr ] − Xr2 =0 • E[Xr (Xt − Xs )|Fs ] = E[Xr Xt − Xr Xs |Fs ] = E[Xr Xt |Fs ] − E[Xr Xs |Fs ] = X r E[Xt |Fs ] − X r Xs =0 • E[(Xt − Xs )(Xs − Xr )|Fs ] = E[Xt Xs − Xs2 − Xt Xr + Xs Xr |Fs ] = E[Xt Xs |Fs ] − E[Xs2 |Fs ] − E[Xt Xr |Fs ] + E[Xs Xr |Fs ] = Xs E[Xt |Fs ] − Xs2 − Xr E[Xt |Fs ] + Xs Xr = Xs2 − Xs2 − Xr Xs + Xs Xr =0 • E[Xr (Xs − Xr )] = E[E[Xr (Xs − Xr )|Fr ]] = E[0] =0 E[Xr (Xt − Xs )] = E[E[Xr (Xt − Xs )|Fs ]] = E[0] =0 E[(Xt − Xs )(Xs − Xr )] = E[E[(Xt − Xs )(Xs − Xr )|Fs ]] = E[0] =0 34 MARTINGALES 5 MARTINGALES n Definition: L Bounded A set of r.vs C is Ln bounded if supX∈C E[|X|n ] < ∞ Theorem: ∞ 2 Every L2 bounded martingale {Xn }∞ n=1 w.r.t. {Fn }n=1 converges in L 2 2 i.e. ∃I ∈ L s.t. limn→∞ E[|Xn − Y | ] = 0 Proof: L2 is complete hence every cauchy sequence converges n−1 X E[(Xk+1 − Xk )2 ] = k=1 n−1 X 2 E[(Xk+1 + Xk2 − 2Xk Xk+1 ] k=1 = n−1 X 2 E[(Xk+1 + Xk2 − 2Xk Xk+1 ] k=1 = E[(Xn − X1 )2 ] by a telescoping sum, orthogonality of increments since Xn is L2 bounded ≤ 2c this yields lim supm≥n≥k E[(Xm − Xn )2 ] = lim supm≥n≥k n→∞ k→∞ ≤ lim supm≥n≥k k→∞ m−1 X E[(Xl+1 − Xl )2 ] l=n ∞ X E[(Xl+1 − Xl )2 ] l=k =0 Remark: 2 The previous theorem S∞ also gives us that Y ∈ L (Ω, F∞ , P) where F∞ := σ ( k=1 Fk ) Theorem: Levy Let X ∈ L2 (Ω, F, P) and {Fk }∞ k=1 a filtration define Xn := E[X|Fn ] then Xn → E[X|F∞ ] in L2 Proof: We have that Xn converges in L2 so let Xn → Y We want to show that E[Y 1A ] = E[X1A ] ∀A ∈ Fn let A ∈ Fn and m ≥ n then notice that E[Xm 1A ] = E[E[X|Fm ]1A ] = E[X1A ] since Xm → Y in L2 we have that lim |E[Xm 1A − Y 1A ]| ≤ lim (E[Xm − Y ]2 1A )1/2 m→∞ by the Cauchy-Schwarz inequality m→∞ =0 hence indeed we have that E[X1A ] = E[Y 1A ] ∀A ∈ Fn We need this ∀A ∈ F∞ 35 5 MARTINGALES define D := {A ∈ F∞ : E[X1A ] = E[Y 1A ]} We want to show that D is a dynkin system. Ω ∈ D holds trivially since Ω ∈ Fn ∀n Let A ∈ D E[Y 1Ac ] = E[Y 1Ω ] − E[Y 1A ] = E[X1Ω ] − E[X1A ] = E[X1Ac ] as required. S∞ Suppose {Ai }∞ i=1 ∈ D are disjoint and A := i=1 Ai " E[Y 1A ] = E Y lim n→∞ " = lim E Y n→∞ = lim n→∞ = lim n→∞ n X k=1 n X # 1Ak # 1Ak by DCT k=1 n X k=1 n X E[Y 1Ak ] E[X1Ak ] k=1 " = lim E X n→∞ " = E X lim n→∞ n X k=1 n X # 1Ak # 1Ak k=1 = E[X1A ] SinceSD is a Dynkin system if I ⊂ D is a π-system then σ(I) ⊂ D ∞ I := n=1 Fn ⊂ D is a π-system since if {Ai }ni=1 ∈ I then ∀i ∃ki s.t. Ai ∈ Fki hence since finite ∃K = maxi {ki } then since {Fn }∞ n=1 is a filtration Tn Ai ∈ FK ∀i hence since FK is a σ-algebra i=1 Ai ∈ FK thus F∞ = σ(I) ⊆ D ⊆ F∞ Lemma: DOOB ∞ 2 ∞ Let {Fn }∞ n=1 be a filtration of (Ω, F, P) and {Xn }n=1 ∈ L an adapted sequence of {Fn }n=1 then 1 ∞ 1 • ∃{An }∞ n=1 ⊂ L and martingale {Mn }n=1 ⊂ L with – A1 = 0 – An+1 Fn -measurable – Xn = Mn + An which is a P-a.s. unique decomposition • For Xn = Mn + An above we have that ∞ {Xn }∞ n=1 is a sub-martingale iff {An }n=1 is increasing P-a.s. 36 5 MARTINGALES Proof: • If we construct the sequence recursively then the sequence will be P-a.s unique A1 = 0 =⇒ M1 = X1 Xn+1 = Mn+1 + An+1 E[Xn+1 |Fn ] = E[Mn+1 |Fn ] + E[An+1 |Fn ] = Mn + An+1 since An+1 An+1 = E[Xn+1 |Fn ] − Mn is clearly Fn measurable. E[Mn+1 |Fn ] = E[Xn+1 − An+1 |Fn ] = E[Xn+1 |Fn ] − E[An+1 |Fn ] = An+1 + Mn − An+1 = Mn hence {Mn }∞ n=1 is a martingale. • An+1 − An = E[Xn+1 |Fn ] − Mn − An = E[Xn+1 |Fn ] − Xn which is positive iff Xn ≤ E[Xn+1 |Fn ] i.e. {Xn }∞ n=1 is a sub-martingale. Definition: Stopping Time Let {Fn }∞ n=1 be a filtration on (Ω, F, P) and let Y : Ω → N ∪ {∞} be a random variable then T is a stopping time if {T ≤ n} ∈ Fn ∀n Lemma: Let (Ω, F, P) be a probability space with stopping times T, S on filtration {Fn }∞ n=1 then T ∩ S := min(T, S), T ∪ S := max(T, S) are stopping times. Proof: Clearly T ∩ S, T ∪ S are r.vs {T ∩ S ≤ n} = {T ≤ n} ∪ {S ≤ n} {T ∪ S ≤ n} = {T ≤ n} ∩ {S ≤ n} Since T, S are stopping times {T ≤ n}, {S ≤ n} ∈ Fn hence {T ∩ S ≤ n}, {T ∪ S ≤ n} ∈ Fn Lemma: Let (Ω, F, P) be a probability space with stopping time T on filtration {Fn }∞ n=1 then the set of events which are observable up top time T FT := {A ∈ F : A ∩ {T ≤ n} ∈ Fn ∀n} is a σ-algebra. Lemma: Let (Ω, F, P) be a probability space with stopping time T on filtration {Fn }∞ n=1 ∞ If {Xn }∞ n=1 is adapted to {Fn }n=1 then 37 Fn -measurable 5 ( XT (ω) (ω) T (ω) < ∞ XT (ω) := 0 o/w is FT measurable Proof: Let Xn : (Ω, F) → (Ω0 , F 0 ) and let A ∈ F 0 {XT ∈ A} = {0 ∈ A, T = ∞} ∪ ∞ [ {Xn ∈ A} ∩ {T = n} ∈ F n=1 hence XT is a random variable Moreover {XT ∈ A} ∩ {T = n} = {Xn ∈ A} ∩ {T = n} ∈ Fn since Xn adapted and T is a stopping time hence {XT ∈ A} ∈ FT Lemma: ∞ Let {Xn }∞ n=1 be a martingale on {Fn }n=1 and T stopping time s.t. T ≤ m a.s. then E[Xm |T ] = XT P-a.s. Proof: XT is FT measurable for k ∈ N|[1,m] and A ∈ FT E[XT 1A 1T =k ] = E[Xk 1A∩{T =k} ] = E[E[Xm |Fk ]1A∩{T =k} ] = E[E[Xm 1A∩{T =k} |Fk ]] = E[Xm 1A∩{T =k} ] m X E[XT 1A ] = E[XT 1A 1T =k ] = k=1 m X E[Xm 1A 1T =k ] k=1 = E[Xm 1A ] hence XT = E[Xm |FT ] Lemma: Let S ≤ T be stopping times on {Fn }∞ n=1 then FS ⊆ FT Proposition: Optimal Stopping Let S ≤ T be two bounded stopping times on {Fn }∞ n=1 and {Xn }∞ n=1 be a martingale on the same filtration. then E[XT |FS ] = XS P-a.s. Proof: T ≤ m hence E[XT |FS ] = E[E[Xm |FT ]|FS ] = E[Xm |FS ] = XS 38 MARTINGALES 5 Proposition: Let S ≤ T be P-a.s. finite stopping times and {Xn }∞ n=1 a sub-martingale then E[XT |FS ] ≥ XS Proposition: Let S ≤ T be P-a.s. finite stopping times and {Xn }∞ n=1 a super-martingale then E[XT |FS ] ≤ XS Theorem: First DOOB Inequality Let {Xn }∞ n=1 be a sub-martingale, X n := maxi≤n Xi and a > 0 then aP(X n ≥ a) ≤ E[Xn 1{X n ≥a} ] Proof: Let T = inf {n ∈ N : Xn ≥ a} which is a stopping time. Then {X n ≥ a} = {T ≤ n} = {XT ∩n ≥ a} hence aP(X n ≥ a) = aP(XT ∩n ≥ a) = aE[1T ∩n ] = E[a1T ∩n ] ≤ E[XT ∩n 1T ∩n ] ≤ E[E[Xn |FT ∩n ]1T ∩n ] = E[E[Xn 1T ∩n |FT ∩n ]] = E[Xn 1T ∩n ] Corollary: If {Xn }∞ n=1 is a non-negative sub-martingale then ∀a ≥ 0, p ≥ 1 ap P(X n ≥ a) ≤ E[Xnp ] Theorem: Second DOOB Inequality If {Xn }∞ n=1 is a non-negative sub-martingale and p > 1 then p p p E[X n ] ≤ E[Xnp ] p−1 Definition: Crosses The sequence {Xn }∞ n=1 crosses the interval [a, b] if ∃i, j s.t. Xi < a < b < Xj Lemma: Let {Xn }∞ n=1 which crosses any rational interval only finitely often. Then {Xn }∞ n=1 converges in R ∪ {±∞} Proof: Suppose liminfn Xn < limsupn Xn then ∃a, b ∈ Q s.t.liminfn Xn < a < b < limsupn Xn then {Xn }∞ n=1 crosses [a, b] i.o. Definition: Up-Crossing ∞ If {Xn }∞ n=1 is an adapted sequence of {Fn }n=1 and a < b then define T0 := 0 and recursively Sk := inf {n ≥ Tk−1 : Xn ≤ a} Tk := inf {n ≥ Sk : Xn ≥ b} 39 MARTINGALES 5 k {Xn }Tn=S k Then is the kth up-crossing of [a, b] Remark: We write the number of up-crossings of [a, b] by time n as Una,b := n X 1Tk ≤n k=1 Lemma: Let {Xn }∞ n=1 be a martingale then E[Una,b ] ≤ Proof: Let Z := ∞ X E[(Xn − a)− ] b−a (XTk ∩n − XSk ∩n ) k=1 then E[(XTk ∩n − XSk ∩n )] = E[XTk ∩n ] − E[XSk ∩n ] = E[E[XTk ∩n |FSk ∩n ]] − E[XSk ∩n ] = E[XSk ∩n ] − E[XSk ∩n ] =0 E[Z] = 0 Suppose Una,b = m then Z ≥ m(b − a) + (Xn − XSm +1∩n ) since Z crosses [a, b] m times and there could be one uncompleted crossing. Xn − XSm +1∩n ≥ Xn − a hence Z≥ ∞ X (1{Una,b =m} m(b − a) + 1{Una,b =m} (Xn − a)) m=0 0 = E[Z] ∞ X ≥ P(Una,b = m)m(b − a) − E[(Xn − a)− ] m=0 = E[Una,b ](b − a) − E[(Xn − a)− ] hence we have the required result. Theorem: Martingale Convergence Theorem Let {Xn }∞ n=1 be a martingale s.t. supn E[(Xn )− ] < ∞ then X∞ := limn→∞ Xn exists P-a.s. moreover X∞ is F∞ measurable and X∞ ∈ L1 Proof: By the previous lemma E[(Xn − a)− ] <∞ E[Una,b ] ≤ b−a by monotone convergence of expectations we have a,b E[U∞ ]<∞ a,b a,b but E[U∞ ] ≥ 0 hence E[U∞ ] < ∞ P-a.s. 40 MARTINGALES 5 MARTINGALES Since a, b ∈ Q there are only a countable number of intervals hence ∃N ∈ F s.t. P(N ) = 0 and ∀a < b ∈ Q, ∀ω ∈ Ω \ N a,b we have that E[U∞ ]<∞ moreover limn→∞ Xn (ω) exists on R ∪ {±∞} ∀ω ∈ Ω \ N ∞ Y := limn→∞ Xn is F∞ since {Xn }∞ n=1 is adapted to {Fn }n=1 E[Y+ ] = E[liminf (Xn )+ ] ≤ liminf E[(Xn )+ ] by Fatou’s lemma ≤ E[(X0 )+ ] by Jensen’s inequality <∞ E[Y− ] = E[liminf (Xn )− ] ≤ liminf E[(Xn )− ] by Fatou’s lemma ≤ E[(X0 )− ] by Jensen’s inequality <∞ hence E[Y ] < ∞ therefore Y ∈ L1 Remark: • The upcrossing lemma and martingale convergence theorem remain true for super martingales • In general we do not have L1 convergence Theorem: Convergence For Uniformly Integrable Martingales Suppose that {Xn }∞ n=1 is a uniformly integrable super-martingale. Then Xn converges P-a.s. and in L1 (Ω, F, P) to some r.v. X∞ ∈ L1 Ω, F∞ , P) and E[X∞ |Fn ] ≤ Xn Proof: For a super-martingale {Xn }∞ n=1 by uniform integrability Z limk→∞ supn |Xn |dP = 0 {|Xn |>k} Z Z (Xn )− dP ≤ |Xn |dP Z Z = |Xn |dP + |Xn |dP {|X |>k} {|Xn |≤k} Z n Z ≤ |Xn |dP + k dP {|Xn |>k} {|Xn |≤k} Z ≤ supn |Xn |dP + kP(|Xn | ≤ k) {|Xn |>k} <∞ Then by the martingale convergence theorem limn→∞ Xn = X∞ P-a.s. and since a.s. convergence and uniform integrability ensure L1 convergence we have convergence in L1 (Ω, F, P) i.e. limn→∞ ||X∞ − Xn || = 0 41 5 MARTINGALES ||E[X∞ |Fm ] − E[Xn |Fm ]|| = ||E[X∞ − Xn |Fm ]|| ≤ ||X∞ − Xm || lim ||E[X∞ |Fm ] − E[Xn |Fm ]|| = 0 m→∞ Corollary: Suppose that {Xn }∞ n=1 is a uniformly integrable sub-martingale. Then Xn converges P-a.s. and in L1 (Ω, F, P) to some r.v. X∞ ∈ L1 (Ω, F∞ , P) and E[X∞ |Fn ] ≥ Xn Corollary: Suppose that {Xn }∞ n=1 is a uniformly integrable martingale. Then Xn converges P-a.s. and in L1 (Ω, F, P) to some r.v. X∞ ∈ L1 (Ω, F∞ , P) and E[X∞ |Fn ] = Xn Theorem: Backwards Martingale Theorem Let (Ω, F, P) be a probability space and {G−n : n ∈ N} sub-σ-algebras of F s.t. \ G∞ := G−k ⊆ G−(n+1) ⊆ G−n ⊆ G−1 k∈N For X ∈ L1 (Ω, F, P) define X−n := E[X|G−n ] 1 then limn→∞ X−n exists P-a.s. and in L Theorem: Strong Law Of Large Numbers Pn Let {Xi }∞ i=1 be i.i.d r.vs with E[|Xn |] < ∞, µ = E[Xi ] and Sn = i=1 Xi then Sn /n → µ P-a.s. Proof: T∞ Define G−n := σ({Si }∞ i=n ) and G−∞ = n=1 G−n then E[X1 |G−n ] = E[X1 |Sn ] = Sn /n ∞ since G−n = σ(Sn , {Xi }∞ i=n+1 ) and independence of {Xi }i=1 hence Sn L := lim n→∞ n exists P-a.s. and in L1 moreover Pn Pn+k Xi Xi L = limsupn i=1 = limsupn i=k n n is measurable w.r.t. τk = σ({Xi }∞ ) Ti=k i hence is measurable w.r.t. τ = k=1 nf tyFk hence by Komogorov’s zero-one law ∃c s.t. P(L = c) = 0 moreover µ := E[Sn /n] and limn→∞ E[Sn /n] = E[L] = c hence indeed µ must be the limit. Theorem: Central Limit Theorem 2 Let {Xk }∞ k=1 ∈ L be i.i.d random variables denote µ := E[Xi ] ν := V ar[Xi ] then n X Xk − µ √ → N (0, 1) Sn := νn k=1 weakly Proof: √ WLOG let µ = 0, ν = 1 since otherwise we can use the transformation X = µ + νZ where Z ∼ N (0, 1) then n n X Xk − µ X Zk √ √ = νn n k=1 k=1 42 5 MARTINGALES hence indeed it suffices to show for µ = 0, ν = 1 We need to show that ∀f ∈ Cb (R) that Z lim E[f (Sn )] = n→∞ ex 2 /2 √ f (x) dx 2π however any f ∈ Cb (R) can be approximated by some f ∈ Cb (R) which has bounded and uniformly continuous first and second derivatives. ∞ Let {Yk }∞ k=1 ∼i,i,d N (0, 1) be independent of {Xk }k=1 then it suffices to show that for n X Y √k Tn := n k=1 we have that denote Xk,n and define lim E[f (Sn ) − f (Tn )] = 0 √ √ = Xk / n. Yk,n = Yk / n n→∞ Wk,n = k−1 X Yj,n + j=1 Then f (Sn ) − f (Tn ) = n X n X Xj,n j=k+1 f (Wk,n + Xk,n ) − f (Wk,n + Yk,n ) k=1 by a telescoping sum, hence by Taylor’s theorem we have 1 2 + RXk,n f (Wk,n + Xk,n ) = f (Wk,n ) + f 0 (Wk,n )Xk,n + f 00 (Wk,n )Xk,n 2 1 2 + RYk,n f (Wk,n + Yk,n ) = f (Wk,n ) + f 0 (Wk,n )Yk,n + f 00 (Wk,n )Yk,n 2 1 2 2 − Yk,n ) + RXk,n − RYk,n f (Wk,n + Xk,n ) − f (Wk,n + Yk,n ) = f 0 (Wk,n )(Xk,n − Yk,n ) + f 00 (Wk,n )(Xk,n 2 E[f (Wk,n + Xk,n ) − f (Wk,n + Yk,n )] ≤ 2 2 E[|f 0 (Wk,n )(Xk,n − Yk,n )|] + E[|f 00 (Wk,n )(Xk,n − Yk,n )/2|] + E[|RXk,n | + |RYk,n |] also from Taylor’s theorem we have 2 |RXk,n | ≤ Xk,n ||f 00 ||∞ so by uniform continuity we have that ∀ε > 0 ∃δ > 0 s.t. 2 |RXk,n | ≤ Xk,n ε ∀|Xk,n | < δ thus 2 2 |RXk,n | ≤ Xk,n ε1{|Xk,n |<δ} + Xk,n ||f 00 ||∞ 1{|Xk,n |≥δ} 43 5 MARTINGALES moreover " n # X E |RXk,n | + |RYk,n | k=1 ≤ n X 2 2 E[Xk,n (ε1{|Xk,n |<δ} + ||f 00 ||∞ 1{|Xk,n |≥δ} ) + Yk,n (ε1{|Yk,n |<δ} + ||f 00 ||∞ 1{|Yk,n |≥δ} )] k=1 n 1X E[Xn2 (ε1{|Xk,n |<δ} + ||f 00 ||∞ 1{|Xk,n |≥δ} ) + Yn2 (ε1{|Yk,n |<δ} + ||f 00 ||∞ 1{|Yk,n |≥δ} )] = n k=1 ! n n X 1 X 2 00 2 2 00 E[Xn ||f ||∞ 1{|Xk,n |≥δ} ] + ≤ 2εE[X1 ] + E[Yn ||f ||∞ 1{|Yk,n |≥δ} ] n k=1 = 2εE[X12 ] + k=1 2E[X12 ||f 00 ||∞ 1{|Xn |≥δ√n} ] →n→∞ 0 By the dominated convergence theorem. t remains to show that the first two terms are null E[f 0 (Wk,n )Xk,n ] = E[f 0 (Wk,n )]E[Xk,n ] by independence =0 0 E[f (Wk,n )Yk,n ] = E[f 0 (Wk,n )]E[Yk,n ] by independence =0 E[f 00 2 (Wk,n )(Xk,n − 2 Yk,n )] 2 2 = E[f 00 (Wk,n )]E[Xk,n − Yk,n ] 1 1 − = E[f 00 (Wk,n )] n n =0 hence indeed the theorem holds. 44